Question

Historically, the average waiting time spent on hold when phoning customer service for a particular bank was 50 seconds. To determine whether the average time people were waiting had increased since they cut the number of customer service staff, the bank recorded the waiting time for a random sample of 15 customers. The results are in the Y column (in seconds)

Y	X1	X2
28	12.6	134
43	11.4	126
45	11.5	143
49	11.1	152
57	10.4	143
68	9.6	147
74	9.8	128
81	8.4	119
82	8.8	130
86	8.9	135
101	8.1	141
112	7.6	123
114	7.8	121
119	7.4	129
124	6.4	135

. Assume that the test is performed at the 5% level of significance and that the distribution of waiting times is approximately normally distributed.

1. State the direction of the alternative hypothesis used to test whether average waiting time had increased. Type gt (greater than), ge (greater than or equal to), lt (less than), le (less than or equal to) or ne (not equal to) as appropriate in the box.

2. Calculate the test statistic correct to three decimal places (hint: use Descriptive Statistics to calculated the standard deviation and sample mean).

3. Use KaddStat to determine the p-value for the test correct to four decimal places.

4. Is the null hypothesis rejected for this test? Type yes or no.

5. Regardless of your answer for 4, if the null hypothesis was not rejected, could we conclude that the average waiting time has increased at the 5% level of significance? yes or no.

Answer 1

1) Since we are testing that if there is increase in the average customer waiting time after cutting the number of service staff so Alternative hypothesis should be average waiting is greater than 50 Hence Use "gt" in the box Ha: average waiting time >50 While HO: average waiting time=50 2) From given that we can easily calculate following values Sample size=n=15 sample.mean = y = 4 y n 1183 = 78.867 15 E (:- )_13067.73 = 30.552 sample.SD = S=1 V n-1 V 14 Now test statistics is calculated as t=] - 50 S/Vn 78.867 – 50 -= 3.659 30.552/ 15 Wheret ist distribution with df=n-1=15-1=14

3) Since test is right tailed test hence P value is given by P(t>3.659)=1-P (t<3.659)=1-0.9987=0.0013 (Using add stat for T distribution with df=14) 4) Since P-value is less than level of significance (0.05) hence we reject the null hypothesis Hence Answer is YES 5) No, since our alternative hypothesis was "average waiting time is more than 50" so if HO is not rejected then how can we accept alternative hypothesis hence Answer is "NO"